Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $a = \dfrac{z - 3}{z^2 + 2z - 15} \div \dfrac{z + 10}{-6z - 30} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{z - 3}{z^2 + 2z - 15} \times \dfrac{-6z - 30}{z + 10} $ First factor the quadratic. $a = \dfrac{z - 3}{(z + 5)(z - 3)} \times \dfrac{-6z - 30}{z + 10} $ Then factor out any other terms. $a = \dfrac{z - 3}{(z + 5)(z - 3)} \times \dfrac{-6(z + 5)}{z + 10} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ (z - 3) \times -6(z + 5) } { (z + 5)(z - 3) \times (z + 10) } $ $a = \dfrac{ -6(z - 3)(z + 5)}{ (z + 5)(z - 3)(z + 10)} $ Notice that $(z - 3)$ and $(z + 5)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ -6(z - 3)\cancel{(z + 5)}}{ \cancel{(z + 5)}(z - 3)(z + 10)} $ We are dividing by $z + 5$ , so $z + 5 \neq 0$ Therefore, $z \neq -5$ $a = \dfrac{ -6\cancel{(z - 3)}\cancel{(z + 5)}}{ \cancel{(z + 5)}\cancel{(z - 3)}(z + 10)} $ We are dividing by $z - 3$ , so $z - 3 \neq 0$ Therefore, $z \neq 3$ $a = \dfrac{-6}{z + 10} ; \space z \neq -5 ; \space z \neq 3 $